Answer: 0.1457
Step-by-step explanation:
Let p be the population proportion.
Given: The proportion of Americans who are afraid to fly is 0.10.
i.e. p= 0.10
Sample size : n= 1100
Sample proportion of Americans who are afraid to fly =
We assume that the population is normally distributed
Now, the probability that the sample proportion is more than 0.11:
![P(\hat{p}>0.11)=P(\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}>\dfrac{0.11-0.10}{\sqrt{\dfrac{0.10(0.90)}{1100}}})\\\\=P(z>\dfrac{0.01}{0.0090453})\ \ \ [\because z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}} ]\\\\=P(z>1.1055)\\\\=1-P(z\leq1.055)\\\\=1-0.8543=0.1457\ \ \ [\text{using z-table}]](https://tex.z-dn.net/?f=P%28%5Chat%7Bp%7D%3E0.11%29%3DP%28%5Cdfrac%7B%5Chat%7Bp%7D-p%7D%7B%5Csqrt%7B%5Cdfrac%7Bp%281-p%29%7D%7Bn%7D%7D%7D%3E%5Cdfrac%7B0.11-0.10%7D%7B%5Csqrt%7B%5Cdfrac%7B0.10%280.90%29%7D%7B1100%7D%7D%7D%29%5C%5C%5C%5C%3DP%28z%3E%5Cdfrac%7B0.01%7D%7B0.0090453%7D%29%5C%20%5C%20%5C%20%5B%5Cbecause%20z%3D%5Cdfrac%7B%5Chat%7Bp%7D-p%7D%7B%5Csqrt%7B%5Cdfrac%7Bp%281-p%29%7D%7Bn%7D%7D%7D%20%5D%5C%5C%5C%5C%3DP%28z%3E1.1055%29%5C%5C%5C%5C%3D1-P%28z%5Cleq1.055%29%5C%5C%5C%5C%3D1-0.8543%3D0.1457%5C%20%5C%20%5C%20%5B%5Ctext%7Busing%20z-table%7D%5D)
Hence, the probability that the sample proportion is more than 0.11 = 0.1457
29% = 0.29
Multiply 1.89 with 0.29
1.89 x 0.29 = 0.5481
Note that it is a markdown, so subtract 0.55 (rounded) with 1.89
1.89 - 0.55 = 1.34
The spark plug now costs ~ $1.34
hope this helps
Let 
, so that 
:
Now the ODE is separable, and we have
Integrating both sides gives
For the integral on the left, rewrite the integrand as
Then
and so
Given that 
, we find
so that the particular solution to this IVP is
Answer:6.05cm
Step-by-step explanation:
Circumference=2×π×r
38=2×22/7×r
38×7=2×22×r
266=44r
r=266/44
r=6.05cm
